Calculate the pH when 10.0 mL of 0.150 M KOH is mixed with 20.0 mL of 0.300 M HBrO (Ka = 2.5 * 10-9

Calculate the pH when 10.0 mL of 0.150 M KOH is mixed with 20.0 mL of 0.300 M HBrO (Ka = 2.5 * 10-9

The Correct Answer and Explanation is:

To calculate the pH of the solution after mixing 10.0 mL of 0.150 M KOH and 20.0 mL of 0.300 M HBrO, we need to understand the neutralization reaction and the subsequent equilibrium in the solution.

Step 1: Write the neutralization reaction

KOH (a strong base) reacts with HBrO (a weak acid) in a 1:1 ratio:KOH(aq)+HBrO(aq)→KBrO(aq)+H2O(l)\text{KOH} (aq) + \text{HBrO} (aq) \rightarrow \text{KBrO} (aq) + \text{H2O} (l)KOH(aq)+HBrO(aq)→KBrO(aq)+H2O(l)

KOH dissociates completely in water, while HBrO only partially dissociates due to its weak acidic nature.

Step 2: Calculate the moles of KOH and HBrO

• Moles of KOH:
  The concentration of KOH is 0.150 M, and the volume is 10.0 mL = 0.010 L.

moles of KOH=0.150 M×0.010 L=0.0015 moles\text{moles of KOH} = 0.150 \, \text{M} \times 0.010 \, \text{L} = 0.0015 \, \text{moles}moles of KOH=0.150M×0.010L=0.0015moles

• Moles of HBrO:
  The concentration of HBrO is 0.300 M, and the volume is 20.0 mL = 0.020 L.

moles of HBrO=0.300 M×0.020 L=0.0060 moles\text{moles of HBrO} = 0.300 \, \text{M} \times 0.020 \, \text{L} = 0.0060 \, \text{moles}moles of HBrO=0.300M×0.020L=0.0060moles

Step 3: Determine the limiting reagent and excess

Since KOH reacts in a 1:1 ratio with HBrO, we can see that KOH is the limiting reagent (0.0015 moles of KOH vs 0.0060 moles of HBrO). After neutralization, we will have:

• 0.0015 moles of KOH reacting with 0.0015 moles of HBrO, leaving 0.0045 moles of HBrO unreacted.

Step 4: Calculate the final concentration of HBrO

The total volume of the mixture is:Total volume=10.0 mL+20.0 mL=30.0 mL=0.030 L\text{Total volume} = 10.0 \, \text{mL} + 20.0 \, \text{mL} = 30.0 \, \text{mL} = 0.030 \, \text{L}Total volume=10.0mL+20.0mL=30.0mL=0.030L

The concentration of remaining HBrO is:[HBrO]=0.0045 moles0.030 L=0.150 M[\text{HBrO}] = \frac{0.0045 \, \text{moles}}{0.030 \, \text{L}} = 0.150 \, \text{M}[HBrO]=0.030L0.0045moles​=0.150M

Step 5: Set up the equilibrium expression for HBrO dissociation

Since HBrO is a weak acid, it will partially dissociate:HBrO⇌H++BrO−\text{HBrO} \rightleftharpoons \text{H}^+ + \text{BrO}^-HBrO⇌H++BrO−

The equilibrium constant KaK_aKa​ for HBrO is given as 2.5×10−92.5 \times 10^{-9}2.5×10−9.

We use the ICE (Initial, Change, Equilibrium) method to calculate the concentration of H+H^+H+ ions.

• Initial concentrations:

[HBrO]=0.150 M,[H+]=0,[BrO−]=0[\text{HBrO}] = 0.150 \, \text{M}, \quad [\text{H}^+] = 0, \quad [\text{BrO}^-] = 0[HBrO]=0.150M,[H+]=0,[BrO−]=0

• Change:

[HBrO] decreases by x,[H+] increases by x,[BrO−] increases by x[\text{HBrO}] \text{ decreases by } x, \quad [\text{H}^+] \text{ increases by } x, \quad [\text{BrO}^-] \text{ increases by } x[HBrO] decreases by x,[H+] increases by x,[BrO−] increases by x

• Equilibrium concentrations:

[HBrO]=0.150−x,[H+]=x,[BrO−]=x[\text{HBrO}] = 0.150 – x, \quad [\text{H}^+] = x, \quad [\text{BrO}^-] = x[HBrO]=0.150−x,[H+]=x,[BrO−]=x

Step 6: Write the equilibrium expression

The expression for the dissociation of HBrO is:Ka=[H+][BrO−][HBrO]=x⋅x0.150−x=2.5×10−9K_a = \frac{[\text{H}^+][\text{BrO}^-]}{[\text{HBrO}]} = \frac{x \cdot x}{0.150 – x} = 2.5 \times 10^{-9}Ka​=[HBrO][H+][BrO−]​=0.150−xx⋅x​=2.5×10−9

Simplifying the equation:x20.150−x=2.5×10−9\frac{x^2}{0.150 – x} = 2.5 \times 10^{-9}0.150−xx2​=2.5×10−9

Since xxx will be very small compared to 0.150 M, we can approximate 0.150−x≈0.1500.150 – x \approx 0.1500.150−x≈0.150, so the equation becomes:x20.150=2.5×10−9\frac{x^2}{0.150} = 2.5 \times 10^{-9}0.150×2​=2.5×10−9

Solving for xxx:x2=(2.5×10−9)×0.150=3.75×10−10x^2 = (2.5 \times 10^{-9}) \times 0.150 = 3.75 \times 10^{-10}x2=(2.5×10−9)×0.150=3.75×10−10x=3.75×10−10=6.12×10−5 Mx = \sqrt{3.75 \times 10^{-10}} = 6.12 \times 10^{-5} \, \text{M}x=3.75×10−10​=6.12×10−5M

Thus, the concentration of H+H^+H+ is 6.12×10−5 M6.12 \times 10^{-5} \, \text{M}6.12×10−5M.

Step 7: Calculate the pH

Finally, we calculate the pH using:pH=−log⁡[H+]=−log⁡(6.12×10−5)=4.21\text{pH} = -\log[\text{H}^+] = -\log(6.12 \times 10^{-5}) = 4.21pH=−log[H+]=−log(6.12×10−5)=4.21

Final Answer:

The pH of the solution is 4.21.